What is the $"molality"$ of a $30g$ mass of sodium hydroxide dissolved in a $100g$ mass of water?

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Answer 1 · 2017-06-06T18:07:59

$"Molality"=7.5*mol*kg^-1$

$"Molality"="Moles of solute"/"Kilograms of solute"$

And here we have $30*g$ $NaOH$ solute..........

So $"Molality"=((30*g)/(40.00*g*mol^-1))/(100*gxx10^-3*g*kg^-1)=7.5*mol*kg^-1.$

What is the "molality""molality" of a 30*g30*g mass of sodium hydroxide dissolved in a 100*g100*g mass of water?